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Let $f$ be a holomorphic function on the open unit disc.

Is $(\sup \vert f \vert)^2 = \sup (\vert f\vert^2)$?

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2 Answers 2

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I don't think this has anything to do with $f$ being a holomorphic function on the open unit disk. The square of the supremum of any set of non-negative numbers is the supremum of the squares.

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For weakly positive real numbers, the inequalities $x \le y$ and $x^2 \le y^2$ are equivalent. Since the modulus of the function is always weakly positive, this is all you need.

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non-negative, to be precise... –  joriki Nov 13 '11 at 19:03
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@joriki: Sorry, I am living in a weakly positive country, will correct. –  Phira Nov 13 '11 at 19:05
    
@joriki: I refuse to use "non-negative" and "non-positive" (or non-decreasing, etc), it goes against human cognitive functioning to gratuitously add negations. –  Phira Nov 13 '11 at 19:08
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Interesting. For my cognitive functioning, "non-negative" is a lot easier to parse than "weakly positive". –  joriki Nov 13 '11 at 19:13
    
@joriki You should try it with a directed concept where you did not already "grow up" with the term. –  Phira Nov 13 '11 at 21:49

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